• [CF498C] Array and Operations


    [CF498C] Array and Operations - 数论,最大流

    Description

    有一个长度为 n 的数组 a 和 m 对数 ((i_1,j_1),(i_2,j_2)...,(i_m,j_m)),对于每对数都满足 (i_k + j_k) 是一个奇数,且每个数都在 1 到 n 之间。你每次操作可需要挑一对数(给定的 m 对里面)(i_k,j_k),然后使 (a[i_k]=frac{a[i_k]}{v},a[j_k]=frac{a[j_k]}{v}),v 是一个不等于 1 的正整数,且 (v)(a[i])(a[j]) 的公约数。问最多可以进行多少次操作。

    Solution

    每次消去一个质因子一定是最优解

    奇数的性质显然保证了每次操作发生在两个不同的部分之间,这是一个显然的二分图

    对于每个数的每个质因数建点,到源/汇(这取决于它位置的奇偶性)的容量为质因子的指数

    对于一个操作 ((i_k,j_k)) 我们在这两个数的所有相同的质因子之间连边,容量可以设为无穷

    #include <bits/stdc++.h>
    using namespace std;
    #define int long long
    
    #ifndef __FLOW_HPP__
    #define __FLOW_HPP__
    
    // v1.1     feat. edge query for Maxf
    
    #include <bits/stdc++.h>
    using namespace std;
    #define int long long
    
    namespace flowsolution
    {
        const int N = 100005;
        const int M = 1000005;
        const int inf = 1e+12;
    
        struct MaxflowSolution
        {
            int *dis, ans, cnt = 1, s, t, *pre, *next, *head, *val;
    
            MaxflowSolution()
            {
                cnt = 1;
                dis = new int[N];
                pre = new int[M];
                next = new int[M];
                head = new int[N];
                val = new int[M];
                fill(dis, dis + N, 0);
                fill(pre, pre + M, 0);
                fill(next, next + M, 0);
                fill(head, head + N, 0);
                fill(val, val + M, 0);
            }
    
            ~MaxflowSolution()
            {
                delete[] dis;
                delete[] pre;
                delete[] next;
                delete[] head;
                delete[] val;
            }
    
            std::queue<int> q;
            int make(int x, int y, int z)
            {
                // cerr << "make " << x << " " << y << " " << z << endl;
                pre[++cnt] = y, next[cnt] = head[x], head[x] = cnt, val[cnt] = z;
                int ret = cnt;
                pre[++cnt] = x, next[cnt] = head[y], head[y] = cnt;
                return ret;
            }
    
            int get_value(int x)
            {
                return val[x];
            }
    
            bool bfs()
            {
                fill(dis, dis + N, 0);
                q.push(s), dis[s] = 1;
                while (!q.empty())
                {
                    int x = q.front();
                    q.pop();
                    for (int i = head[x]; i; i = next[i])
                        if (!dis[pre[i]] && val[i])
                            dis[pre[i]] = dis[x] + 1, q.push(pre[i]);
                }
                return dis[t];
            }
    
            int dfs(int x, int flow)
            {
                if (x == t || !flow)
                    return flow;
                int f = flow;
                for (int i = head[x]; i; i = next[i])
                    if (val[i] && dis[pre[i]] > dis[x])
                    {
                        int y = dfs(pre[i], min(val[i], f));
                        f -= y, val[i] -= y, val[i ^ 1] += y;
                        if (!f)
                            return flow;
                    }
                if (f == flow)
                    dis[x] = -1;
                return flow - f;
            }
    
            int solve(int _s, int _t)
            {
                s = _s;
                t = _t;
                ans = 0;
                for (; bfs(); ans += dfs(s, inf))
                    ;
                return ans;
            }
        };
    
        struct CostflowSolution
        {
            struct Edge
            {
                int p = 0, c = 0, w = 0, next = -1;
            } * e;
            int s, t, tans, ans, cost, ind, *bus, qhead = 0, qtail = -1, *qu, *vis, *dist;
    
            CostflowSolution()
            {
                e = new Edge[M];
                qu = new int[M];
                bus = new int[N];
                vis = new int[N];
                dist = new int[N];
                fill(qu, qu + M, 0);
                fill(bus, bus + N, -1);
                fill(vis, vis + N, 0);
                fill(dist, dist + N, 0);
                ind = 0;
            }
    
            ~CostflowSolution()
            {
                delete[] e;
                delete[] qu;
                delete[] vis;
                delete[] dist;
            }
    
            void graph_link(int p, int q, int c, int w)
            {
                e[ind].p = q;
                e[ind].c = c;
                e[ind].w = w;
                e[ind].next = bus[p];
                bus[p] = ind;
                ++ind;
            }
    
            void make(int p, int q, int c, int w)
            {
                graph_link(p, q, c, w);
                graph_link(q, p, 0, -w);
            }
    
            int dinic_spfa()
            {
                qhead = 0;
                qtail = -1;
                fill(vis, vis + N, 0);
                fill(dist, dist + N, inf);
                vis[s] = 1;
                dist[s] = 0;
                qu[++qtail] = s;
                while (qtail >= qhead)
                {
                    int p = qu[qhead++];
                    vis[p] = 0;
                    for (int i = bus[p]; i != -1; i = e[i].next)
                        if (dist[e[i].p] > dist[p] + e[i].w && e[i].c > 0)
                        {
                            dist[e[i].p] = dist[p] + e[i].w;
                            if (vis[e[i].p] == 0)
                                vis[e[i].p] = 1, qu[++qtail] = e[i].p;
                        }
                }
                return dist[t] < inf;
            }
    
            int dinic_dfs(int p, int lim)
            {
                if (p == t)
                    return lim;
                vis[p] = 1;
                int ret = 0;
                for (int i = bus[p]; i != -1; i = e[i].next)
                {
                    int q = e[i].p;
                    if (e[i].c > 0 && dist[q] == dist[p] + e[i].w && vis[q] == 0)
                    {
                        int res = dinic_dfs(q, min(lim, e[i].c));
                        cost += res * e[i].w;
                        e[i].c -= res;
                        e[i ^ 1].c += res;
                        ret += res;
                        lim -= res;
                        if (lim == 0)
                            break;
                    }
                }
                return ret;
            }
    
            pair<int, int> solve(int _s, int _t)
            {
                s = _s;
                t = _t;
                ans = 0;
                cost = 0;
                while (dinic_spfa())
                {
                    fill(vis, vis + N, 0);
                    ans += dinic_dfs(s, inf);
                }
                return make_pair(ans, cost);
            }
        };
    } // namespace flowsolution
    
    #endif
    
    int n, m, a[105];
    map<int, int> mp[105];
    
    map<int, int> factorize(int x)
    {
        map<int, int> ans;
        int lim = sqrt(x);
        for (int i = 2; i <= lim; i++)
        {
            while (x % i == 0)
            {
                x /= i;
                ans[i]++;
            }
        }
        if (x > 1)
            ans[x]++;
        return ans;
    }
    
    map<pair<int, int>, int> node;
    
    signed main()
    {
        ios::sync_with_stdio(false);
    
        cin >> n >> m;
        for (int i = 1; i <= n; i++)
            cin >> a[i], mp[i] = factorize(a[i]);
    
        flowsolution::MaxflowSolution flow;
        int ind = 2;
        int S = 1, T = 2;
    
        for (int i = 1; i <= n; i++)
        {
            for (auto [x, y] : mp[i])
            {
                node[{i, x}] = ++ind;
                if (i & 1)
                    flow.make(S, ind, y);
                else
                    flow.make(ind, T, y);
            }
        }
    
        while (m--)
        {
            int i, j;
            cin >> i >> j;
            for (auto [x, y] : mp[i])
                if (node[{j, x}])
                {
                    if (j & 1)
                        swap(i, j);
                    flow.make(node[{i, x}], node[{j, x}], 1e9);
                }
        }
    
        cout << flow.solve(S, T) << endl;
    }
    
  • 相关阅读:
    HashMap,Hash优化与高效散列
    Dubbo Overview
    模板引擎 引自 《PHP核心技术与最佳实践》
    使用 phpStorm 开发
    使用 Zend_Studio 开发
    Symfony 2.0 认识Request, Response, Session, Cookie
    Symfony 建立一个Bundle
    Symfony 从路由认识它
    信鸽推送.net 服务端代码
    c#输出json,其中包含子json (可以含 无限级 子json)的方法思路
  • 原文地址:https://www.cnblogs.com/mollnn/p/14358803.html
Copyright © 2020-2023  润新知